Exercise 5.4.5 (Equation $x^2 + y^2 = 3(u^2 + v^2)$)

Proof. Suppose for the sake of contradiction that $x^2+y^2=3(u^2+v^2)$, where $(x,y,u,v)\ne (0,0,0,0)$. Then $d=x\wedge y\wedge z\wedge t\ne 0$. If $X=x∕d,Y=y∕d,U=u∕d,V=v∕d$, we obtain after simplification by $d^2$,

Then $X^2+Y^2\equiv 0(\mathit{mod}3)$. Since $X^2\equiv 0(\mathit{mod}3)$ or $X^2\equiv 1(\mathit{mod}3)$, this congruence implies

Then $3^2\mid X^2+Y^2=3(U^2+V^2)$, thus $3\mid U^2+V^2$, os $U^2+V^2\equiv 0(\mathit{mod}3)$. By the same reasoning,

Therefore $3\mid X\wedge Y\wedge U\wedge V=1$. This is a contradiction, which proves that $x^2+y^2=3(u^2+v^2)$ has no solution $(x,y,u,v)\ne (0,0,0,0)$. □